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Pay attention to these six points for capacitor step-down, no need to worry about analyzing circuit!

by:Shenmao     2021-04-30
Pay attention to the following six points for capacitor step-down: 1. Choose an appropriate capacitor according to the current size of the load and the working frequency of the alternating current, rather than the voltage and power of the load. 2. Current-limiting capacitors must be non-polar capacitors, and electrolytic capacitors must not be used. And the withstand voltage of the capacitor must be above 400V. The most ideal capacitor is an oil-immersed iron-case capacitor. 3. Capacitor step-down cannot be used in high-power conditions because it is not safe. 4. Capacitor step-down is not suitable for dynamic load conditions. 5. Similarly, capacitor step-down is not suitable for capacitive and inductive loads. 6. When DC work is required, half-wave rectification should be used as much as possible. Bridge rectification is not recommended. And to meet the condition of constant load. After paying attention to these six points, I will simply cite a few circuits as examples, so that everyone can understand more quickly, and master the six points of capacitor step-down in the future, and there is no need to worry about analyzing the circuit! 1. Circuit 1: This type of circuit is usually used to obtain non-isolated low-current power supplies at low cost. Its output voltage is usually a few volts to three dozens of volts, depending on the Zener regulator tube used. The amount of current that can be provided is proportional to the capacity of the current-limiting capacitor. When half-wave rectification is used, the current (average value) that can be obtained per microfarad capacitance is: (International Standard Unit) I(AV)u003d0.44*V/Zcu003d0.44*220*2*Pi*f*Cu003d0.44 *220*2*3.14*50*Cu003d30000Cu003d30000*0.000001u003d0.03Au003d30mA If full-wave rectification is used, double current (average value) can be obtained: I(AV)u003d0.89*V/Zcu003d0.89 *220*2*Pi*f*Cu003d0.89*220*2*3.14*50*Cu003d60000Cu003d60000*0.000001u003d0.06Au003d60mA Generally, the full-wave rectifier of this type of circuit has a slightly larger current, but it is due to floating Ground, stability and safety are worse than the half-wave rectifier type, so it uses less. When using this circuit, you need to pay attention to the following: 1. It is not isolated from 220V AC high voltage, please pay attention to safety and prevent electric shock! 2. The current-limiting capacitor must be connected to the live wire, the withstand voltage must be large enough (greater than 400V), and a series of anti-surge impact, insurance resistance and parallel discharge resistance should be added. 3. Pay attention to the power consumption of the Zener tube, and it is strictly forbidden to disconnect the Zener tube. 2. Circuit 2: The simplest capacitor step-down DC power supply circuit and its equivalent circuit are shown in Figure 1. C1 is a step-down capacitor, generally 0.33~3.3uF. Assuming C1u003d2uF, its capacitive reactance XCLu003d1/(2PI*fC1)u003d1592. Since the on-resistance of the rectifier tube is only a few ohms, the dynamic resistance of the voltage regulator tube VS is about 10 ohms, the current limiting resistance R1 and load resistance RL are generally 100~200, and the filter capacitor is generally 100uF~1000uF, and its capacitive reactance is very It is small and can be ignored. If R represents the equivalent resistance of all components except C1, the AC equivalent circuit of the figure can be drawn. At the same time, the condition of XC1>R is satisfied, so the voltage vector can be drawn. Since R is much smaller than XC1, the voltage drop VR on R is also much smaller than the voltage drop on C1, so VC1 is approximately equal to the power supply voltage V, that is, VC1u003dV. According to the principle of electrical engineering, the relationship between the average value Id of the rectified DC current and the average value I of the alternating current is Idu003dV/XC1. If C1 is in uF, then Id is in milliamperes. For 22V, 50 Hz alternating current, Idu003d0.62C1 can be obtained. From this, the following two conclusions can be drawn: (1) When the power transformer is used as a rectifier power supply, when the various parameters in the circuit are determined, the output voltage is constant, and the output current Id changes with the increase or decrease of the load; ( 2) When using a capacitor step-down as a rectifier circuit, since Idu003d0.62C1, it can be seen that Id is proportional to C1, that is, after C1 is determined, the output current Id is constant, but the output DC voltage is different with the size of the load resistance RL Change within a certain range. The smaller the RL, the lower the output voltage, and the larger the RL, the higher the output voltage. The value of C1 should be selected according to the load current. For example, the load circuit requires a 9V working voltage and the average load current is 75 mA. Since Idu003d0.62C1, it can be calculated as C1u003d1.2uF. Taking into account the loss of the voltage regulator tube VD5, C1 can be 1.5uF, at this time the actual current provided by the power supply is Idu003d93 mA. The voltage regulation value of the voltage regulator tube should be equal to the working voltage of the load circuit, and the selection of its stable current is also very important. Since the capacitor step-down power supply provides a constant current, which is similar to a constant current source, it is generally not afraid of load short-circuiting, but when the load is completely open, the entire 93 mA current will pass through the R1 and VD5 loops, so the maximum stability of VD5 The current should be 100 mA. Since RL and VD5 are connected in parallel, while ensuring that RL uses 75 mA working current, there is still 18 mA current flowing through VD5, so the minimum stable current must not be greater than 18 mA, otherwise the voltage stabilization effect will be lost. The value of the current-limiting resistor cannot be too large, otherwise it will increase the power loss and increase the withstand voltage requirement of C2.
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